3 years ago

Prove: AAB

Mathematics

1 answer:

goldenfox [79]3 years ago

6 0

Hmmm… I wish I could help you with this one, sorry

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Please help me ! Reduced to lowest terms

Mandarinka [93]

Answer:

Step-by-step explanation:

7) 2ab + 4ac = 2*a*b + 2 *2 *a*c

= 2a(b + 2c)

$\dfrac{2ab^{2}c}{2ab+4ac}=\dfrac{2*a* b^{2}*c}{2a(b+2c)}\\\\\\\text{2a in the denominator and numerator get cancelled}\\\\ = \dfrac{b^{2}c}{b+2c}$

8) ac - a²c² = c( a - a²c)

bc - abc = c(b - ab)

$\dfrac{ac - a^{2}c^{2}}{bc-abc^{2}}=\dfrac{c(a-a^{2}c)}{c(b-ab)}\\\\$

$=\dfrac{a-a^{2}c}{b-abc}$

9) b²+ 4b - 5 = b² + 5b - 1b - 5

= b(b + 5) - 1(b + 5)

= (b +5)(b - 1)

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= b(b + 5) + 3(b + 5)

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2 years ago

PLEASE DONT SKIP<br><br><br> show all work

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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